Question

A mass block of mass m1 is attached to the rigid and weightless bar ABC whose other end is pin-connected to the wall The bar is supported by a spring of spring constant of k3 at its midpoint B. AB BC-a-1m. Another block of mass m2 is connected to the first block by a spring of spring constant k1 and is connected to the fixed ground by a spring of spring constant k2. The size of both blocks are ignored. Assume that the system is originally at its static equilibrium position and the motion starts from rest. Only small vertical motion is considered 71 x (t) 71 k-k,-8N/mk-32N/m 2 Q1: Find the stiffness matrix and the mass matrix of the system Q2: Find the natural frequencies and the corresponding modal shape vectors of the system by solving an eigenvalue problem. Solve the problem manually first and then use MATLAB eig-function to verify the result. Q3: What is the modal matrix of the system? Check its orthogonality with respect to the system stiffness matrix and the mass matrix, respectively. What are the generalized mass and stiffness coefficients of the system? Q4: Use the energy approach to modify the system mass matrix if the bar mass in taken into account. Assume that the bar is uniform and its mass M-3kg. What are the modified natural frequency values?

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Answer #1

Since the su Shem !6 orgi no po si hm, we do nor to condan rawiy 2a since ㈨.UHAo displacemants k, (x1-x2) 2久 fVt vn D amd D

Let the natwral reanemies be co Then 2- 16-4 w 16-4, (4) Mode s 2. lo- 02 ㄥ Mode

Command Window >> K= [16-8, -8 16]; >> [modeShape , naturalFreq2] eig (MK) = modeShape -0.7071 -0.7071 -0.7071 0.7071 natural

to generate he eigen pair. [vne de Shape, nahna/ Freg 2] = eig (MXK) We aet mode Shape 2- the Mmarix vwhich is bosi M comn. 2

Command Window >> K = [16-8, -8 16]; J5 ans - 16 0 48 ans -

Here,X KX Let be a cams fore ordi nat such ak 3.CE) Pre mutti ply boh sdes by XT ana get 거2/nae ke generaな242 mans coed-CLeh

,707 四..70 小.te , we take X w a chwiu live nue· saw.efnal.ve nLt 2 3 2 Phe fi (e Lagranges er(ua.hm ǐら PS)-..24-: f.lt)う4..2

So,

K = \begin{bmatrix} 16 & -8\\ -8 & 16 \end{bmatrix} \ M = \begin{bmatrix} 5 & 0\\ 0 & 4 \end{bmatrix}

Let, the modified natural frequencies be \Omega . So,

\begin{vmatrix} 16 - 5\Omega ^2 & -8\\ -8 & 16-4\Omega ^2 \end{vmatrix} =0 \Rightarrow (16-5\Omega ^2)(16-4\Omega ^2)-64=0

\Rightarrow 5\Omega ^4 - 36 \Omega ^2 + 48 =0

\Rightarrow \Omega ^2 = \frac{18 \pm 2\sqrt{21}}{5}

\Rightarrow \mathbf{\Omega = 1.33 \ rad/s, \ 2.33 \ rad/s}

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